(2) In Figure B, rotate the straight line AB counterclockwise around point B for a certain angle and intersect with the straight line CD at point Q, as shown in Figure C, then what is the quantitative relationship between ∠ bpd, ∠ b, ∠ d and ∠ bqd? (No proof required)
(3) According to the conclusion of (2), find the degree of ∠A+∠B+∠C+∠D+∠E+∠F in Figure D.
answer
1) is invalid. The conclusion is ∠ BPD = ∠ B+∠ D.
Extend the BP intersection CD to point e,
∫AB∨CD
∴∠B=∠BED
And ≈BPD =∠BED+∠D,
∴∠BPD=∠B+∠D.
(2) Conclusion: ∠ BPD = ∠ BQD+∠ B+∠ D.
(3) connect EG and expand it,
According to the exterior angle property of triangle ∠AGB=∠A+∠B+∠E,
∠∠AGB =∠CGF，
In quadrilateral CDFG, ∠ CGF+∠ C+∠ D+∠ F = 360,
∴∠A+∠B+∠C+∠D+∠E+∠F=360。
2
As shown in the figure, point A and point B are on straight lines CM and DN, respectively, CM∑DN.
(1) As shown in figure 1, if AB is connected, then ∠CAB+∠ABD=
(2) As shown in Figure 2, the point P 1 is a point in the straight line CM and DN, connecting AP 1 and BP 1. Verification: ∠ cap1+∠ AP1b+∠ p1BD.
(3) As shown in Figure 3, points P 1 and P2 are points in the straight lines CM and DN connecting P 1, P 1P2 and P2B .. Try to find ∠ cap1∠ AP1P2+∠.
(4) According to the above rules, guess and write the degree of ∠ CAP 1+∠ AP 1p2+ ... ∠ P5bd directly (without writing the flow).
answer
( 1)∵CM∨DN。
∴∠cab+∠abd= 180；
(2) the point P 1 is a parallel line parallel to CM and DN,
∴∠ap 1e+∠cap 1= 180，∠EP 1 b+∠p 1bd = 180，
∴∠cap 1+∠ap 1b+∠p 1bd=∠ap 1e+∠cab+∠ep 1b+∠p 1bd= 180+ 180 = 360；
(3) the intersections P 1 and P2 are parallel lines parallel to CM and DN,
∴∠ap 1e+∠cap 1= 180，∠EP 1p 2+∠p 1p2f = 180，∠FP2B+∠P2BD= 180，
∴∠cap 1+∠ap 1p2+∠p 1p2b+∠p2bd=∠ap 1e+∠cap 1+∠ep 1p2+∠p 1p2f+∠fp2b+∠p2bd=3× 180 = 540；
(4)∠cap 1+∠AP 1p 2+……∠P5BD = 6× 180 = 1080。
3 Calculation: If the cube root of 5x+ 19 is 4, find the square root of 2x+ 18.
answer
Solution: According to the meaning of the question: 5x+ 19=43,
I.e. 5x=45,
Then x=9,
Then 2x+ 18=36,
Then the square root of 2x+ 18 is 6.
It is known that the square root of 2a- 1 is 3, and the arithmetic square root of 3a+b- 1 is 4. Find the cube root of 12a+2b.
answer
Solution: The square root of ∵2a- 1 is 3.
∴ 2a- 1 = (3) 2 and a = 5；;
The arithmetic square root of 3a+b- 1 is 4.
∴3a+b- 1= 16, substitute a=5, 3×5+b- 1= 16, and get b=2.
∴ 12a+2b= 12×5+4=64，
∴
3 64
=4,
That is, the cube root of 12a+2b is 4.
five
As shown in figure 1, in the plane rectangular coordinate system, two points A and B start from the origin O at the same time, with point A moving in the positive direction of X axis at m unit lengths per second and point B moving in the positive direction of Y axis at n unit lengths per second.
(1) When the known motion is 1 sec, point B moves more than point A 1 unit; When moving for 2 seconds, the sum of the distances between point B and point A is 6 units. Find m and n;
(2) As shown in Figure 2, assuming that the bisector of the adjacent complementary angle of ∠OBA and the bisector of the adjacent complementary angle of ∠OAB intersect at point P, does the size of ∠P change? If it is unchanged, find its value; If there are any changes, explain the reasons.
(3) If the bisector of ∠OBA and the inverse extension of the bisector of ∠OAB intersect at point Q, does the size of ∠Q change? If it is unchanged, find its value; If there are any changes, please explain the reasons.
Solution to the answer: (1)∵ When the movement is known as 1 sec, point B moves 1 unit more than point A, so n-m =1; When moving for 2 seconds, the sum of the distances between point B and point A is 6 units Ⅷ.
n？ m= 1
2n+2m=6
Solution:
n=2
m= 1
(2) the size of ∠ p remains the same, ∠ p = 45.
∵∠ Oba+∠ OAB =180-∠ O = 90; The sum of adjacent complementary angles of ∠ Oba and ∠OAB is 180-∠ Oba+(180-∠ OAB) = 360-90 = 270;
∫BP bisects the adjacent complementary angle of∠ ∠OBA, and PA bisects the adjacent complementary angle of∠ ∠OAB.
∴∠PBA+∠PAB= 135
∠∠PBA+∠PAB+∠P = 180
∴∠p= 180-(∠PBA+∠PAB)= 180- 135 = 45；
(3) The size of ∠ q remains the same, and ∠ q = 45.
∫∠BAX is the external angle of △AOB.
∴∠BAX=∠O+∠OBA
∵BQ shares∠ ∠BAO, and shares∠ ∠BAX.
∴∠ 1=∠2,∠3=∠4
∴∠3=
1
2
(∠O+∠OBA)=45 +∠2
∫∠3 is the outer corner of △ABQ.
∴∠3=∠Q+∠2
∴∠Q=∠3-∠2=45 +∠2-∠2=45。
six
As shown in figure 1, in the plane rectangular coordinate system, △AOB is a right triangle, ∠ AOB = 90, and the hypotenuse AB intersects the Y axis at point C. 。
(1) If ∠A=∠AOC, verification: ∠ B = ∠ BOC;
(2) As shown in Figure 2, extend the intersection of AB and X axis at point E and let O be OD⊥AB. If ∠DOB=∠EOB and ∠A=∠E, find the degree of ∠A;
(3) As shown in Figure 3, the extension line of the bisector where ∠AOM and ∠BCO intersect with FO is at point P, ∠ A = 40. When △ABO rotates around point O (hypotenuse AB and positive half axis of Y axis always intersect at point C), does the degree of △ P change? If it does not change, find its degree; If yes, please explain why.
answer
(1) proves that ∵△AOB is a right triangle,
∴∠A+∠B=90，∠AOC+∠BOC=90，
∠∠A =∠AOC，
∴∠b=∠boc；
Solution: (2)≈A+∠ABO = 90, ∠ DOB+∠ ABO = 90,
∴∠A=∠DOB，
And ∠∠DOB =∠EOB, ∠A=∠E,
∴∠DOB=∠EOB=∠OAE=∠OEA，
∵∠DOB+∠EOB+∠OEA=90，
∴∠a=30；
(3) The number of ∠ p remains the same, ∠ P = 25. The reasons are as follows: (If you only answer the same question, you won't score)
∠∠AOM = 90-∠AOC，∠BCO=∠A+∠AOC，
There are also: equal division ∠AOM, CP equal division ∠BCO,
∴∠FOM=45 -
1
2
∠AOC ①，∠PCO=
1
2
∠A+
1
2
∠AOC ②，
①+②: ∠ PCO+∠ FOM = 45+
1
2
∠A，
∴∠P= 180 -(∠PCO+∠FOM+90)
= 180 -(45 +
1
2
∠A+90)
= 180 -(45 +20 +90 )
=25 .
Xiangxi is famous as "the hometown of ponkan". On a Sunday during the ponkan harvest season, Qingshan Middle School sent some students from Grade 8 (1) and (2) to the orchard to help villagers pick ponkan, including two boys and eight girls from Grade 8 (1). Eight grade (2) class transferred four male students and six female students, and * * * picked 880 Jin of mandarin oranges. How many kilograms did each boy pick on average? How many Jin of mandarin oranges do female students pick on average?
Answer: Suppose that male students each pick an average of X kilograms of ponkan, and female students each pick an average of Y kilograms of ponkan.
From the meaning of the question, get
2x+8y=840
4x+6y=880
Get a solution
x= 100
y=80
.
A: On average, boys pick 100 Jin of ponkan, while girls pick 80 Jin of ponkan.
A new teaching building has been built in a middle school in Guangzhou. The building has four access doors, two of which are as big as the main entrance, and two side doors. During the security check, four doors were tested: when a main entrance and two side doors were opened at the same time, 280 students could pass through every minute; When a main entrance and a side door are opened at the same time, 200 students can pass through every minute.
(1) How many students can a main entrance and a side door pass through every minute on average?
(2) In an emergency, because students are crowded, the efficiency of going out will be reduced by 20%. It is now stipulated that in case of emergency, students in the whole building should be evacuated safely through these four doors within five minutes. Suppose there are 32 classrooms in this teaching building, with a maximum of 45 students in each classroom. Q: Do these four doors meet the requirements? Please explain the reason.
answer
(1) On average, one main entrance passes through X students and one side entrance passes through Y students every minute.
According to the meaning of the question, you must:
x+2y=280
x+y=200
Solution:
x= 120
y=80
A: On average, a main entrance can pass 120 students per minute, and a side door can pass 80 students.
(2) The maximum number of students in this building is 32×45= 1440 (name).
When crowded, the number of students who can pass through the four gates within 5 minutes is 5× 2 (120+80) (1-20%) =1600 (name).
∵ 1600> 1440,
∴ The four doors built meet the safety requirements.
In order to reward the outstanding students in the art performance, the class teacher sent Xiao Liang, a life member, to the stationery store to buy prizes for the winning students. Xiao Liang found that if he bought 1 notebook and 3 pens, he needed 18 yuan. Buying two notebooks and five pens costs 3 1 yuan.
(1) How much do you want for a notebook and a pen?
(2) The class fee paid by the head teacher to Xiao Liang is 100 yuan, and there are 24 students who need to be rewarded (each person will be rewarded with a prize). If the number of pens purchased is not less than the number of notebooks, what are Xiao Liang's purchase plans?
Solution: (1) Let each notebook be X yuan and each pen be Y yuan.
According to the meaning of the question:
x+3y= 18
2x+5y=3 1
Solution:
x=3
y=5
A: Set up a 3 yuan for each notebook and a 5 yuan for each pen.
(2) Buy a (24-m) pen if you buy M's notebook.
According to the meaning of the question:
3m+5(24？ m)≤ 100
m≤24？ m
Solution: 12≥m≥ 10.
M is a positive integer.
∴m= 10 or 1 1 or 12.
There are three purchase schemes: ① purchase 10 notebook and 14 notebook.
② If you buy 1 1 notebook, you will buy 13 pens.
③ Buy 12 notebooks and buy 12 pens.
10 A shopping mall bought two kinds of goods, A and B, for 36,000 yuan, and after the sale, * * * made a profit of 6,000 yuan. The purchase price of each commodity is 120 yuan, and the selling price is 138 yuan. B Purchase price 100 Yuan, selling price 120 yuan.
(1) How many A and B goods did the mall buy?
(2) For the second time, the shopping mall bought two kinds of goods, A and B, at the original purchase price. The number of goods purchased by B remained unchanged, while the number of goods purchased by A was twice that of the first time. The goods sold by A are sold at the original price, while the goods sold by B are sold at a discount. If you sell these two kinds of goods, you need to make a profit of not less than 8 160 yuan for the second business activity. What is the lowest price of B's goods?
Answer (1) Set up a shopping mall and buy X pieces of Class A goods and Y pieces of Class B goods.
120x+ 100y=36000
( 138? 120)x+( 120？ 100)y=6000
Solution:
x=200
y= 120
.
A: The mall bought 200 pieces of Class A goods, 120 pieces of Class B goods.
(2) Assuming that the price of each commodity of commodity B is Z yuan, according to the meaning of the question, you can get
120(z- 100)+2×200×( 138- 120)≥8 160，
Solution: z ≥ 108.
A: The lowest selling price of commodity B is 108 yuan per piece.
Ok!!